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If x and y are allowed to be complex, then there are three solutions (if x is non-zero) and so x has three cube roots. A real number has one real cube root and two further cube roots which form a complex conjugate pair. For instance, the cube roots of 1 are:
In number theory, the radical of a positive integer n is defined as the product of the distinct prime numbers dividing n. Each prime factor of n occurs exactly once as a factor of this product: r a d ( n ) = ∏ p ∣ n p prime p {\displaystyle \displaystyle \mathrm {rad} (n)=\prod _{\scriptstyle p\mid n \atop p{\text{ prime}}}p}
Every non-zero number x, real or complex, has n different complex number nth roots. (In the case x is real, this count includes any real nth roots.) The only complex root of 0 is 0. The nth roots of almost all numbers (all integers except the nth powers, and all rationals except the quotients of two nth powers) are irrational. For example,
A solution in radicals or algebraic solution is an expression of a solution of a polynomial equation that is algebraic, that is, relies only on addition, subtraction, multiplication, division, raising to integer powers, and extraction of n th roots (square roots, cube roots, etc.).
For example, taking the statement x + 1 = 0, if x is substituted with 1, this implies 1 + 1 = 2 = 0, which is false, which implies that if x + 1 = 0 then x cannot be 1. If x and y are integers, rationals, or real numbers, then xy = 0 implies x = 0 or y = 0. Consider abc = 0. Then, substituting a for x and bc for y, we learn a = 0 or bc = 0.
The symbol was first seen in print without the vinculum (the horizontal "bar" over the numbers inside the radical symbol) in the year 1525 in Die Coss by Christoff Rudolff, a German mathematician. In 1637 Descartes was the first to unite the German radical sign √ with the vinculum to create the radical symbol in common use today. [3]
Replace x by the average (x + a/x) / 2 between x and a/x. Repeat from step 2, using this average as the new value of x. That is, if an arbitrary guess for is x 0, and x n + 1 = (x n + a/x n) / 2, then each x n is an approximation of which is better for large n than for small n.
In the case of two nested square roots, the following theorem completely solves the problem of denesting. [2]If a and c are rational numbers and c is not the square of a rational number, there are two rational numbers x and y such that + = if and only if is the square of a rational number d.